Complex rate problem

[crackgmat007]

Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?

(1) Train B arrived in New York before Train A arrived in Boston.

(2) The distance between New York and Boston is greater than 140 miles.

OA - D can someone solve this problem? Tx

[mehravikas]

Are you sure that the OA is D??

[raghavsarathy]

IMO - E


Speed of train A = 100mph

They have together travelled for 120 mins

Statement 1

since train B arrived before Train A , and train B leaves at 3 :50 PM , train A has travelled for atleast 50 mins.

Now the remaining 70 mins must be equally between the 2 trains so that tain B reaches before train A. Of this 70 mins , we can say Train A takes 40 mins , train B 30 mins
Or train A 35 mins , train B 25 mins

All satisfying Statement 1. Since we dont obtain unique value ,not sufficient


Statement 2
Given distance d > 140 miles.
We are not told abt the speed of train B. Hence the time taken to reach would vary based on the speed. NOt sufficient.


Combing both the statements.

Train A takes atleast 141/100 i.e 1.41 hrs

For esy calculation lets take it as 1.4 hrs i.e 1 hour 24 mins

Now the remaining 36 mins must be divided between A and B. This can vary based on the speed of B. Hence not sufficient.

[crackgmat007]

yeah, OA is D. I have a very long solution to this question. Was wondering if there is an easier way to solve it.

[vinayakdl]

I agree with D.

Let me know if this makes sense.

x= speed of train B
d= total dist between NY and Boston

so if combined time is 2hrs we get

2 = d/100 +d/x

since they meet 1 hr after A has started (B has travelled for 10 mins...1/6hr)

x/6 + 100(1) = d


Solving both we get
x = 300 or 200
and corresponding d = 150 or 133.33

for 300 speed it takes B 20mins from time of meeting to NY
and for 200 speed it takes 30mins. We are left with additional 50mins

1) now if B arrives earlier it means it had speed of 300 and hence sufficient

2) if d is greater than 140, our other solution 150 so we have speed as 300 and hence 20mins and hence sufficient.

This is one hell of a problem to solve in 2mins...it took me well over 10...

Let me know if i am making any mistakes above.

Vinayak

[crackgmat007]

I agree that this problem cannot be solved quickly. I think your approach is in line with the solution I have. Tx.

[Matt@VeritasPrep]

I hate to resurrect this thread, but I was asked about it today and realized there wasn't an expert answer.

How I'd do it:

Let b = Train B's rate. We're going to solve this problem entirely in terms of this variable.

When the two trains meet, Train B has traveled for (1/6) of an hour at b mph, so Train B's distance = b/6.

After this, Train A has b/6 miles left to travel, and Train B has 100 miles left to travel. Since the trains have traveled for 70 minutes already, they have 50 minutes left to travel, or 5/6 of an hour.

Train B's remaining D = 100 and its R = b, so its T = 100/b.

Train A's remaining D = b/6 and its R = 100, so its T = b/600. But we also know that its remaining T can be written as 5/6 - 100/b, or Total T Left For Both Trains - B's Remaining T.

Since b/600 and 5/6 - 100/b represent the same thing, we can say that

b/600 = 5/6 - 100/b

b = 500 - 60000/b

b² = 500b - 60000

b² - 500b + 60000 = 0

and this quadratic is easy to factor:

(b - 200) * (b - 300) = 0

So B's rate is either 200mph or 300mph.

Remember that the distance for the trip = b/6 + 100 miles, since B travels b/6 before the meeting and 100 miles after it.

With that in mind, if B's rate = 200, then the distance = 200/6 + 100 and B's time for the trip = 2/3 of an hour. If this is the case, B travels 30 minutes after the meeting.

If B's rate = 300, then the distance = 300/6 + 100 and B's time for the trip = 1/2 an hour. If this is the case, B travels 20 minutes after the meeting.

Notice that we STILL HAVEN'T used the statements! We can do virtually all the work without them.

S1 tells us that B travels for less than 25 minutes after the meeting. Only b = 300 from our two solutions above fits, so this is sufficient.

S2 tells us that B travels for less than 26 minutes after the meeting. (If the total distance = 140 miles, then A will take a total of 84 minutes to make the trip, meaning A will travel for 24 minutes after the meeting. But the distance is greater than that, so A will travel > 24 minutes, forcing B to travel < 26 minutes.) Again, only b = 300 fits, and this is also sufficient.